Modulus Algebra (FP2 Complex Numbers)

Modulus algebra is something that is not taught in any of edexcel’s textbooks but is rather assumed knowledge so here’s a little example if you get stuck like me when the mark scheme doesn’t explain the working step by step!

3 = [2i/w -3i]

3 = [(2i – 3iw)/w]

3 = [2i – 3iw]/[w] This means exactly the same as the line above except now you can move [w] to the other side

3[w] = [2i – 3iw]

3[w] = [2 – 3w][i] By looking at an Argand diagram it’s easy to see that [i] = 1

3[w] = [2 – 3w]

3[w] = [3w – 2][-1]

3[w] = [3w – 2]

3[w] = [w – 2/3][3] You could combine this step and the previous step and simply take out [-3] but I did both steps for clarity!

[w] = [w – 2/3] The solution of this is obviously a bisector of the points (0,0) and (0, 2/3) with the equation of the line being u= 1/3

I hope this has helped! If there’s any other maths questions you are stuck on please leave a comment and I’ll do a post on it 🙂

The Square Puzzle

Start with an 8 x 8 grid:

Next, enter the number 1 into any box in the grid of your own choice.

You have now got to move around the grid as a knight piece does in chess; you can only move 2 squares up/down then 1 square left/right or vice versa.

Number consecutively each box you land on.

Here’s an example;

From this you can see I placed the 22 in the wrong box – if I had chosen one of the other 3 possibilities I could have continued.

So, can you fill all 64 boxes?


 From the Daily Post: You have three hundred words to justify the existence of your favorite person, place, or thing. Failure to convince will result in it vanishing without a trace. Go!

For anybody who hasn’t seen any of youtuber Vihart’s videos before, she recently created a series of interesting videos about hexaflexagons; essentially a thin strip of paper folded into a hexagon by first folding it into a line of equilateral triangles, then following the folds in the triangles to make a hexagon.


So what? Well, they’re not just any old hexagon, you can turn them inside out (or flexing) to find different sides, and doing this in a certain way means you disocver new sides. Logic would tell you that it would have two sides, but you can make ones with 3 or even 6 sides.


This feature is why they should exist – they’re a fun way of teaching maths to children because it’s visual and interactive – you can see for yourself the effects of maths in real-time in your hands, instead of on a whiteboard or screen.


Even more useful is their unusual property, with the six sided hexaflexagon, of having the same sides in different states – to understand better what I mean by this video. This allows children to be introduced to diagrams, in this case a Feynman diagram, and helps them to develop their lateral thinking.

Still not convinced hexaflexagons are worth saving? They can also be used to help exaplin A2 Chemistry, in particular the topic of chirality, because depending on how you twisted the original piece of paper, you get a non-superimposable mirror image hexaflexagon – one way the flaps face clockwise, and in the other they face antoclockwise.


Above all, they are fun and can be used to improve artistic ability – what you draw on one face will appear different on another so there begins the challenge of a pattern that looks good on all the faces.

You can even make hexaflexagons out of tortillas for a delicious yet mathematical combination!


I hope I’ve convinced you that hexaflexagons are worth saving, and don’t forget to cast your vote in the poll below to protect the humble hexaflexagon!

Stopping Distances


Instead of learning them off by heart for your theory test, remember that the thinking distance is always the speed (mph) in feet, the braking distance is the speed in feet times a constant, where the constant is 1 plus 0.5 for every 10 mph over 20 mph, then the stopping distance is those two numbers added together – simple really! The method is summarised in the table below:

FP1 – Complex Numbers

What are complex numbers? They’re numbers which are made of a real part, a, and imaginary part, bi, in the form a + bi, note: imaginary and real numbers can never be combined into one term. 

Imaginary number allow you to use the root of a negative number, for example √-4 = 2i.

Adding and subtracting complex numbers is simple; you just treat the real and imaginary parts separately.  

e.g. (2 + 5i) + (7 + 3i) = (2 + 7) + (5 + 3)i = 9 + 8i             

e.g. (6 + 3i) – (4 – 9i) = (6 – 4) + (3 – -9)i = 2 – 6i

Multiplying complex numbers is the same as multiplying in algebra, but beware: as i = √-1, i2 = -1.

e.g. (2 + 3i) x (4 + 5i); using FOIL/parrot’s beak method for multiplying brackets you get 8 + 10i + 12i + 14i2 which equals 8 + 22i – 14  so the final answer is – 6 + 22i.

In order to divide complex numbers, you must first know their complex conjugate. This is simply switching the sign between the real number and imaginary number from + to – or vice versa. Algebraically this means a + bi turns into a – bi. The pair of complex numbers is called the complex conjugate pair. The convention in mathematics is to call one of them z and the other one z*, it doesn’t matter which is which.

To divide two complex numbers, turn them into a fraction, then times both top and bottom by the complex conjugate of the denominator.

e.g. (10 +5i) divided by (1 + 2i) turns into (10 + 5i)/(1 + 2i) x (1 – 2i)/(1 – 2i) which turns into (10 + 5i)(1 – 2i)/(1 + 2i)(1 – 2i) and by using the method learnt for multiplication above it is easy to solve this fraction to get the answer of 4 – 3i. (Some of you may have met this technique before when rationalising denominators, and will see that the denominator of the combined fraction is the difference of 2 squares)

Sometimes you may be given the roots of a quadratic equation, which are always a conjugate pair. This means the equation will be (x – α)(x –β) or x2 – (α + β)x + αβ.

Complex numbers can be shown on a type of graph called an Argand diagram. This is the same as your normal type of graph, but now the x-axis is for the real part of your complex number, and the y-axis is for the imaginary part, and the vectors created by the complex numbers all begin from the origin.

The modulus of a complex number: │x + iy│= √(x2 + y2)

The argument of a complex number (arg z) is the angle θ (usually in radians) between the positive side of the real axis and the vector created by the complex number on an Argand diagram. If z = x + iy, then θ = arctan(y/x).  (Note: if when you draw the vector the complex number makes is to the left of the y (imaginary) axis, take you answer for θ away from 180 (if using degrees) or π (if using radians).). 

Complex numbers can also be written in the form  z = r(cos θ + isin θ) where r is the modulus, and where θ is between -180 and 180 degrees (or between – π and π).

Another property of complex numbers is that if you take any two complex numbers, z1 and z2, then │z1 z2│= │z1││z2│.

You can also find the square root of a complex number. To do this you create simultaneous equations.

e.g. z = 3 + 4i: (a + bi)2 = 3 + 4i then expand to get a2 – b2 + 2abi = 3 + 4i. Then equate the real numbers together in one equation, and the imaginary numbers together in another; a2 – b2 = 3 and 2ab = 4, then solve as you would normal simultaneous equations. In this case the answers are 2 + i and – 2- i.

The techniques you’ve learnt so far can also help you solve cubic or even quartic equations.

Cubic equations: either all 3 roots are real, or one is real and the other 2 roots are a conjugate pair, so to solve it divide the cubic by the real answer, then complete the square to find the conjugate pair.

Quartic equations: either all 4 roots are real, 2 are real and the other 2 are a conjugate pair, or the roots are 2 conjugate pairs.

This is all you need to know about complex numbers – I hope you’ve found it useful! And if there are techniques from GCSE and AS/A2 maths I’ve used or mentioned here that you are unsure about, such as diving cubics by (x + a), please comment and I’ll make another post explaining them – I’ve already completed my A-Level maths this year hence why I’ve started with a current I covered from Further Maths a few weeks ago.